Convert two integers to a long [SOLVED]

Projects Programming
1

As per title, can someone post some code to convert an integer to a long?

Regards

2

Your subject sounds like you want to perform some kind of operation on two integers, with a long result. The content of your post sounds like you want to convert a single integer to a single long.

Can you tell us, specifically, what you really want to do?

3

I have a lsb and a msb part of a 32bit bit variable, therefore two 16bit ints.

I want to combine them back into a single 32bit variable

4

how about

long result;
int high;
int low;

high = 0x1234;
low = 0xabcd;

result = (high << 16) + low;
5

Well that will fail. The default integer size is 16 bits on the Arduino so
any shift-left-by-16 on an int is going to return 0.

copy the high int to the long result, shift the long left by 16, or in the low part...

6

great catch. That probably explains why some of my prior bit manipulation didn't work.

result = high;
result = result  << 16;
result += low;
7

Yes, needs to be casted.

Im trying this but the damm thing cuts anything over 16bit wide.
Just confirmed on AVR freaks the code is OK

   uint32_t number = (((uint32_t)(odo_ms<<16)) + (odo_ls));
   Serial.print((uint32_t)number);

what on earth?????

8

This works. It casts the first argument which is mentioned below by MarkT,

void setup() {
  Serial.begin(9600);
  unsigned int LSB = 0x00ff;
  unsigned int MSB = 0xff00;
  unsigned long Result;
  Result = (unsigned long)MSB << 16 | LSB;
  Serial.println(Result,HEX);
}

void loop() {}
9

casemod:
Yes, needs to be casted.

Im trying this but the damm thing cuts anything over 16bit wide.
Just confirmed on AVR freaks the code is OK

   uint32_t number = (((uint32_t)(odo_ms<<16)) + (odo_ls));

Serial.print((uint32_t)number);




what on earth?????

Because the expression "odo_ms << 16" is int and thus a 16 bit shift is used.
You need to cast odo_ms itself.

10

MarkT:

casemod:
Yes, needs to be casted.

Im trying this but the damm thing cuts anything over 16bit wide.
Just confirmed on AVR freaks the code is OK

   uint32_t number = (((uint32_t)(odo_ms<<16)) + (odo_ls));

Serial.print((uint32_t)number);




what on earth?????

Because the expression "odo_ms << 16" is int and thus a 16 bit shift is used.
You need to cast odo_ms itself.

AHA $)

Of course, create a temp uint32_t var and shift that.

uint32_t MSB = odo_ms;
uint32_t number = ((MSB<<16) + (odo_ls));
11

Or, in your original code, move the cast:

   uint32_t number = ((((uint32_t)odo_ms<<16)) + (odo_ls));

There are way more parentheses in that statement than are needed.

   uint32_t number = ((uint32_t)odo_ms<<16) + odo_ls;